proving this inequality $\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$

Question

proving this inequality $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$$

I tried with $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \int_{\left | z \right |=2}^{} \left | \frac{dz}{z^2+1} \right |$$

but I dont know what next

I think maybe we need to use triangular inequality

Well, the integral can be evaluated quite easily. If you do it correctly, then you will obtain this very trivial inequality $$0\leq\frac{4\pi}{3},.$$ — Batominovski, Sep 16 '18 at 17:13

score 1 · Accepted Answer · answered Sep 04 '13 at 15:45

1

$$\left|\;\;\oint\limits_{|z|=2}\frac{dz}{z^2+1}\;\;\right|\le\oint\limits_{|z|=2}\frac{dz}{|z^2+1|}\stackrel{\text{Est. Lem.}}\le\max_{|z|=2}\left(\frac1{|z|^2-1}\right)\cdot l\left(\{|z|=2\}\right)=$$

$$=\frac13\cdot4\pi=\frac{4\pi}3$$

Est.Lem. = Estimation Lemma of Cauchy

answered Sep 04 '13 at 15:45

DonAntonio

211,718
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do you know another link help me with complexe integral inequality ? – Sep 04 '13 at 15:50
@rama , simply google "Estimation Lemma" . – DonAntonio Sep 04 '13 at 15:52

score 0 · Answer 2 · answered Sep 04 '13 at 15:49

0

Hint: Note that by the triangle inequality $$ |z^{2}+1|\geq||z^{2}|-|1||=|2^{2}-1|=3 $$

since $z$ satisfies $|z|=2$ on the path of the integration.

answered Sep 04 '13 at 15:49

Belgi

23,150

proving this inequality $\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$

2 Answers2